Past exam of the theoretical physics I course of the University of Cambridge / 2017 / 1 / d / Solution

Position-dependent density. The density ${\mathcal{L}}_1$ remains independent of $\psi$ itself, so the shift Noether charge survives with its altered canonical momentum:A prescribed generic function $\rho (z)$ supplies a fixed spatial scale and breaks scale invariance. For the dilation-current candidate with the corresponding ${\mathcal{H}}_1$ and ${\Pi }_1$, the scale-breaking sources in derivative field theories giveThus $Q_2$ is generally not conserved. Constant density recovers the earlier case; specially scale-invariant profiles and suitably invariant domains would require separate treatment.

Quartic strain. The extra term $-\alpha ({\psi }^{\prime }{)}^4$ in ${\mathcal{L}}_2$ again preserves a uniform field shift. Its time momentum is unchanged, while its spatial current becomes $-AK{\psi }^{\prime }-4\alpha ({\psi }^{\prime }{)}^3$. ThereforeThe new Hamiltonian density is ${\mathcal{H}}_2={\mathcal{H}}_0+\alpha ({\psi }^{\prime }{)}^4$. Under waveform dilation, the quartic term scales with $b^4$ rather than $b^2$, andHence there is no corresponding conserved $Q_2$ for nonzero generic $\alpha$, even if the energy in its candidate expression is updated.

Field-dependent quadratic stiffness. In ${\mathcal{L}}_3$, the term $-\beta {\psi }^2({\psi }^{\prime }{)}^2$ breaks the uniform field shift:when spatial flux vanishes. Thus $Q_1$ is not generally conserved. However all terms still contain exactly two derivatives in total, with a field of zero dilation weight. The factor ${\psi }^2$ gains no extra power of $b$, so waveform dilation remains an action symmetry. Its charge isHere $\Pi =A\rho \dot{\psi }$ as before. Relative to the original charge, the energy term changes; the symmetry survives despite the nonlinear field dependence. These statements concern symmetry conservation, separately from any restrictions on parameters needed for positive elastic energy.